Subject: Re: [PATCH 04/21] sched: implement scheduler notifiers
From: Peter Zijlstra <peterz@infradead.org>
To: Tejun Heo <tj@kernel.org>
Cc: linux-kernel@vger.kernel.org, jeff@garzik.org, mingo@elte.hu,
       akpm@linux-foundation.org, jens.axboe@oracle.com, rusty@rustcorp.com.au,
       cl@linux-foundation.org, dhowells@redhat.com, arjan@linux.intel.com,
       torvalds@linux-foundation.org, avi@redhat.com, andi@firstfloor.org,
       fweisbec@gmail.com
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	 <1258391726-30264-5-git-send-email-tj@kernel.org>
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Date: Mon, 16 Nov 2009 19:41:04 +0100
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On Tue, 2009-11-17 at 02:15 +0900, Tejun Heo wrote:
> @@ -1939,6 +1953,8 @@ static int effective_prio(struct task_struct *p)
>   */
>  static void activate_task(struct rq *rq, struct task_struct *p, int wakeup)
>  {
> +       sched_notifier_call(p, SCHED_NOTIFIER_ACTIVATED, activated, wakeup);
> +
>         if (task_contributes_to_load(p))
>                 rq->nr_uninterruptible--;
>  
> @@ -1951,6 +1967,8 @@ static void activate_task(struct rq *rq, struct task_struct *p, int wakeup)
>   */
>  static void deactivate_task(struct rq *rq, struct task_struct *p, int sleep)
>  {
> +       sched_notifier_call(p, SCHED_NOTIFIER_DEACTIVATED, deactivated, sleep);
> +
>         if (task_contributes_to_load(p))
>                 rq->nr_uninterruptible++;
> 

I really hate exposing activate/deactivate.

You say:

> Activated and deactivated are called
> when a task's readiness to run changes.

How is that not clear from the out hook? It would seem to me that when
you get scheduled out with a p->state != TASK_RUNNING you're not ready.
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